Maths Extension 2:Integration

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Integration by Substitution

Examples: For, \int\sqrt{a^2 + x^2}\,dx use the substitution x = tanθ.

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Partial Fractions & Rational Functions

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Handy Partial Fractions Trick

E.g. Convert \frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\, into partial fraction form.

You start off with letting \frac{15x^2-56x+47}{(x-1)(x-2)(x-3)} \equiv \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\,

One way to solve the problem is to multiply both sides to get two polynomials and equate the coefficients and solve a few simultaneous equations. The other way which is not taught in HSC but is useful to know is the Residue Theorem. This is somewhat related to Remainder/Factor theorem.

To use this, you simply multiply a factor in the denominator and then substitue the "zero" into the fraction, so:

  • To find A\,, substitue x=1\, in (x-1)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\, so you get
    • A = \frac{15(1)^2-56(1)+47}{((1)-2)((1)-3)}=\frac{6}{2}=3\,
  • To find B\,, substitue x=2\, in (x-2)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\, so you get
    • B = \frac{15(2)^2-56(2)+47}{((2)-1)((2)-3)}=\frac{-5}{-1}=5\,
  • To find C\,, substitue x=3\, in (x-3)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\, so you get
    • C = \frac{15(3)^2-56(3)+47}{((3)-1)((3)-2)}=\frac{14}{2}=7\,

Hence \frac{15x^2-56x+47}{(x-1)(x-2)(x-3)} \equiv \frac{3}{x-1}+\frac{5}{x-2}+\frac{7}{x-3} \,

This method only works if the factor you are working with looks like x-a\, and not x^n+a\, or (x-a)^n\,.

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Integration by Parts (Reverse Product Rule)

Integration by parts allows the student to perform integration of the form \int f(x)g(x)dx with relative ease.

Students of Mathematics Extension 2 would already be familiar with the rule for differentiating the product of two functions: \frac{d}{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x). Integration by parts uses this fact to construct a rule applicable to integration, as follows:

Suppose G'(x) = g(x); that is, suppose the primitive of g is G. Thus, by the product rule and recasting:

\frac{d}{dx}f(x)G(x)= f'(x)G(x) + f(x)g(x)

f(x)g(x) = \frac{d}{dx}f(x)G(x) - f'(x)G(x)


Integrating both sides gives:

\int f(x)g(x)dx =\int \frac{d}{dx} f(x)G(x) - \int f(x)G(x)dx
=f(x)G(x) - \int f'(x)G(x)dx

which is the rule of integration by parts.

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Trigonometric Integrals


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Reduction Formulae

For example:

Find \int sin^nx\, dx

and hence find \int_{0}^{\frac{\Pi}{4}} sin^9x\, dx

Solution:

Let
\int sin^nx=I_n \,

And let
sin^nx = sin^{n-1}*sinx\,

Integrating by parts
u = sin^nx \,
u'= (n-1)(sin^{n-2}x)(cosx) \,

v'=sinx\,
v = -cosx\,

NOTE: To evaluate u'\,, use the chain rule.

I_n = uv-\int u'v\, dx

I_n = -sin^nxcosx-\int (n-1)(sin^{n-2}x)(cosx)(-cosx)\, dx

Take a factor of (n − 2) out of the integral.

I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x)(cos^2x)\, dx

Using sin^2x+cos^2x=1\, substitute cos^2x=(1-sin^2x)\,


I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x)(1-sin^2x)\, dx

\int (sin^{n-2}x)(1-sin^2x)\, dx = \int (sin^{n-2}x) - \int (sin^nx)\, (by expanding and separating)

I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x) - (n-1)\int (sin^nx)\,

Add (n-1)\int sin^nx\, to both sides.

For the Left side you have:

\int (sin^nx)+n\int (sin^nx)-\int (sin^nx)\,
= n\int (sin^nx)\,

n\int (sin^nx) = -sin^nxcosx+(n-1)\int (sin^{n-2}x)\,dx

Divide by n

\int (sin^nx) = \frac{-sin^nxcosx}{n}+\frac{n-1}{n}\int (sin^{n-2}x)\,dx

This is our general solution for \int sin^nx \,

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Special Properties

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